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An axiomatic approach to block decompositions of rings
Journal of Algebra 284 (2005) 578–592 www.elsevier.com/locate/jalgebra
An axiomatic approach to block decompositions of rings Masaya Matsuura Department of Mathematical Informatics, Graduate School of Information Science and Technology, University of Tokyo, Bunkyo-ku, Tokyo 113-8656, Japan Received 5 February 2004 Available online 13 December 2004 Communicated by Paul Flavell
Abstract In calculations with matrices, block calculations play an important role. To elucidate the essential structures of block decompositions, we shall in this paper introduce a simple system of axioms which guarantees block calculations of rings. The axioms can be interpreted as rules of some kind of information filters. We shall also give another system of axioms in terms of idempotents which is equivalent to the above mentioned system and is similar to the Kolmogorov axioms of probability spaces. 2004 Elsevier Inc. All rights reserved. Keywords: Block decompositions; Block calculations; Idempotents
1. Introduction We assume that each ring treated in this paper has an identity element 1. Let R be any ring and (Ω, F , P ) be any triplet such that Ω is a non-empty set, F is a finitely additive class on Ω and P is a function from F × F to 2R that satisfies (i) P (Ω, Ω) = R, E-mail address: [email protected]. 0021-8693/$ – see front matter 2004 Elsevier Inc. All rights reserved. doi:10.1016/j.jalgebra.2004.10.012
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(ii) P (A, C) + P (B, C) = P (A ∪ B, C), P (A, B) + P (A, C) = P (A, B ∪ C), (iii) P (A, Ω)P (Ω, B) = P (A, B), P (Ω, A)P (Ac , Ω) = P (∅, ∅), for any A, B, C ∈ F . Here the sets S + T and ST (S, T ∈ 2R ) are defined by S + T = {s + t; s ∈ S, t ∈ T },
ST = {st; s ∈ S, t ∈ T }.
If S = ∅ or T = ∅, then we put S + T = ST = ∅. The triplet (Ω, F , P ) can be seen as an information filter with an entrance and an exit. The ring R corresponds to an information set. The first and the second variable of the function P correspond to the entrance and the exit of the filter, respectively. Each element of F indicates the situation of the filter. For example, P (Ω, ∅) indicates the elements of R which can go through this filter when the entrance of the filter is perfectly open and the exit is completely closed. More generally, P (A, B) indicates the information which can pass the filter when the situations of the entrance and the exit are A and B, respectively. Axiom (i) indicates that if both the entrance and the exit are completely open, then all information can go through the filter. We note that in general, even if both the entrance and the exit are completely closed, some information can pass this filter. Axiom (ii) gives a rule of parallel connections of the filter. Axiom (iii) corresponds to series connections. In Section 2, we shall investigate fundamental properties of (Ω, F , P ). In Section 3, we shall see that under an additional condition, each element of R can be divided into block components and block calculations can be applied. Moreover, generalized inverses are examined in connection with (Ω, F , P ). Section 4 is devoted to the study of a special case where the function P is reduced to a one-parameter function. In Section 5, we shall characterize (Ω, F , P ) in terms of idempotents of R. The argument in this paper can be interpreted as reformulation of classical topics on decompositions of rings from a new axiomatic point of view (cf. [3,4]). We conclude this section with two examples. Example 1.1. Let R = ni=1 Ii be a ring such that Ii (i = 1, 2, . . . , n) is a two-sided ideal of R. Then, we can construct (Ω, F , P ) that satisfies Axioms (i)–(iii) as follows: Ω = {1, 2, . . . , n}, P (A, B) = {0} (A ∩ B = ∅),
F = 2Ω ,
P (A, B) =
i∈A∩B
This triplet has the following additional properties: (i) P (∅, ∅) = {0}, (iv) P (A, Ac ) = P (∅, ∅) (A ∈ F ).
Ii
(otherwise).
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Example 1.2. More generally, let R = n+1 i=1 Ii be a ring such that Ii (i = 1, 2, . . . , n + 1) is a right ideal of R. Further, we assume I n+1 is a two-sided ideal of R. Then, there exists an idempotent decomposition 1 = n+1 i=1 ei such that ei ej = 0
(i = j ),
Ii = ei R (1 i n + 1).
We define (Ω, F , P ) as Ω = {1, 2, . . . , n}, F = 2Ω , ei R ei + In+1 (A, B ∈ F ). P (A, B) = i∈A
i∈B
Then, (Ω, F , P ) satisfies Axioms (i)–(iii). Especially, P ({i}, Ω) = Ii ⊕ In+1 (1 i n) and P (∅, ∅) = In+1 . For example, we can show P (A, Ω)P (Ω, B) = P (A, B) as follows. Since In+1 = en+1 R is a two-sided ideal, we have Ren+1 R = en+1 R. Therefore ei Ren+1 = {0}, ei R ei = ei R. i∈A
i∈A
i∈Ω
i∈A
So ei R ei R ei + In+1 = P (A, B). P (A, Ω)P (Ω, B) ⊂ i∈A
i∈Ω
i∈B
To show the converse relation, let x ∈ P (A, B). Then, there exist r ∈ R and s ∈ In+1 such that x= ei r ei + s. i∈A
i∈B
Hence, noting that s ∈ en+1 R, we have ei r ei + en+1 ei · 1 · ei + s = x. i∈A
i∈Ω
i∈Ω
i∈B
This gives P (A, B) ⊂ P (A, Ω)P (Ω, B). Other axioms can be verified in a similar fashion.
2. Fundamental properties Let R be any ring. In this section, we shall investigate fundamental properties of any triplet (Ω, F , P ) satisfying Axioms (i)–(iii). First, we shall see that P (A, B) is a subgroup of R.
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Theorem 2.1. For any A, B ∈ F , P (A, B) is a subgroup of R. Moreover, P (A, Ω) and P (Ω, B) are a right and a left ideal of R, respectively. Proof. Let A, B ∈ F . Then from Axioms (i) and (ii),
P (A, B) + P A, B c + P Ac , B + P Ac , B c = R. Thus, P (A, B) is non-empty. Let x ∈ P (A, B). Then Axiom (iii) shows there exist y ∈ P (A, Ω) and z ∈ P (Ω, B) such that x = yz. Furthermore, Axiom (i) implies that −1 ∈ P (Ω, Ω). Hence from Axiom (iii), we have y · (−1) ∈ P (A, Ω)P (Ω, Ω) = P (A, Ω). So, noting Axiom (iii) again, we have −x = (−y) · z ∈ P (A, Ω)P (Ω, B) = P (A, B). Moreover, Axiom (ii) implies that if x, y ∈ P (A, B), then x + y ∈ P (A, B). Hence P (A, B) is a subgroup. Furthermore, from Axioms (i) and (iii), P (A, Ω)R = P (A, Ω) and RP (Ω, B) = P (Ω, B). This concludes the proof. 2 Theorem 2.2. Let A1 , A2 , B1 , B2 ∈ F satisfy A1 ⊂ A2 and B1 ⊂ B2 . Then, P (A1 , B1 ) ⊂ P (A2 , B2 ), that is, P (A1 , B1 ) is a subgroup of P (A2 , B2 ). Proof. With the help of Axiom (ii), we have P (A1 , B1 ) + P (A2 , B1 ) = P (A2 , B1 ). From Theorem 2.1, 0 ∈ P (A2 , B1 ). Thus P (A1 , B1 ) ⊂ P (A2 , B1 ). Similarly P (A2 , B1 ) ⊂ P (A2 , B2 ). 2 Corollary 2.1. For any A1 , A2 , B1 , B2 ∈ F , we have P (A1 , B1 )P (A2 , B2 ) ⊂ P (A1 , B2 ). Proof. This follows from Theorem 2.2 and Axiom (iii).
2
The next aim is to examine P (∅, ∅). Theorem 2.3. The group P (∅, ∅) satisfies (a) (b) (c) (d)
P (A, ∅) = P (∅, A) = P (∅, ∅) (A ∈ F ), P (∅, ∅)P (∅, ∅) = P (∅, ∅), P (A, B1 )P (B2 , C) = P (∅, ∅) (A, B1 , B2 , C ∈ F , B1 ∩ B2 = ∅), P (∅, ∅) is a two-sided ideal of R.
Proof. Let A ∈ F . From Theorem 2.2, P (∅, ∅) ⊂ P (A, ∅). Further, from Axioms (i), (iii) and Theorem 2.2, P (A, ∅) ⊂ P (A, ∅)P (Ω, Ω) ⊂ P (Ω, ∅)P (Ω, Ω) = P (∅, ∅).
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Thus P (A, ∅) = P (∅, ∅). Similarly, P (∅, A) = P (∅, ∅). Therefore, from Axiom (iii), P (∅, ∅)P (∅, ∅) = P (∅, Ω)P (Ω, ∅) = P (∅, ∅). To show (c), let A, B1 , B2 , C ∈ F and B1 ∩ B2 = ∅. From Theorem 2.2 and Axiom (iii),
P (A, B1 )P (B2 , C) ⊂ P (Ω, B1 )P B1c , Ω = P (∅, ∅). Furthermore, from Theorem 2.2, P (A, B1 )P (B2 , C) ⊃ P (∅, ∅)P (∅, ∅) = P (∅, ∅). Thus we have (c). So, P (∅, ∅)R = P (∅, ∅)P (Ω, Ω) = P (∅, ∅). Similarly, RP (∅, ∅) = P (∅, ∅). This gives (d). 2 Theorem 2.4. For any A1 , A2 , B1 , B2 ∈ F , P (A1 , B1 ) ∩ P (A2 , B2 ) = P (A1 ∩ A2 , B1 ∩ B2 ). In particular, if A1 ∩ A2 = ∅ or B1 ∩ B2 = ∅, then P (A1 , B1 ) ∩ P (A2 , B2 ) = P (∅, ∅). Proof. From Theorem 2.2, we have P (A1 ∩ A2 , B1 ) ∩ P (A1 ∩ A2 , B2 ) ⊂ P (A1 , B1 ) ∩ P (A2 , B2 ). We shall show the converse relation. Let x ∈ P (A1 , B1 ) ∩ P (A2 , B2 ). From Axioms (i) and (ii), 1 and x are divided as 1 = e1 + e2 + e3 , x = x1 + x2 ,
e1 ∈ P (Ω, A1 ∩ A2 ), e2 ∈ P Ω, Ac1 , e3 ∈ P Ω, Ac2 ,
x1 ∈ P (A1 ∩ A2 , B1 ), x2 ∈ P (A1 \ A2 , B1 ).
Then, x = e1 x + (e2 + e3 )x = e1 x1 + e1 x2 + (e2 + e3 )x = x1 − (e2 + e3 )x1 + e1 x2 + (e2 + e3 )x. From Theorem 2.3(c), we know that all terms except the first term of the last line of the above equation belong to P (∅, ∅). Therefore, Theorem 2.2 gives x ∈ P (A1 ∩ A2 , B1 ). Similarly, x ∈ P (A1 ∩ A2 , B2 ). Thus, P (A1 , B1 ) ∩ P (A2 , B2 ) = P (A1 ∩ A2 , B1 ) ∩ P (A1 ∩ A2 , B2 ). Repeating a similar procedure, we obtain P (A1 ∩ A2 , B1 ) ∩ P (A1 ∩ A2 , B2 ) = P (A1 ∩ A2 , B1 ∩ B2 ).
2
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Concerning the quotient ring R/P (∅, ∅), we have Theorem 2.5. Let R be a quotient ring of R defined by R = R/P (∅, ∅). We denote x = x + P (∅, ∅) (x ∈ R). Furthermore, we define a function P : F × F → 2R by P (A, B) = x; x ∈ P (A, B)
(A, B ∈ F ).
Then P is well-defined and has the following properties: (a) The triplet (Ω, F , P ) satisfies Axioms (i)–(iii) for R. (b) Let A, B ∈ F . Then P (A, B) = R if and only if P (A, B) = R. (c) Let A, B ∈ F . Then P (A, B) = {0} if and only if P (A, B) = P (∅, ∅). Especially, P (∅, ∅) = {0}. Proof. Since it is a routine work to show (a)–(c), we only verify that the function P is well-defined. Let x ∈ P (A, B) and x = y. Then from Theorem 2.2, y = x + (y − x) ∈ P (A, B) + P (∅, ∅) = P (A, B). Thus, P is well-defined. 2 At the end of this section, we shall investigate when P (A, B) becomes R. Lemma 2.1. Let x be any regular element of R. For any A ∈ F , if x ∈ P (A, Ω)∪P (Ω, A), then P (A, Ω) = P (Ω, A) = R. Proof. Assume x ∈ P (A, Ω). From Theorem 2.1, P (A, Ω) is a right ideal. Hence, P (A, Ω) = R. We now show P (Ω, A) = R. From Axiom (iii), P (Ω, Ac ) ⊂ P (Ω, Ac ) × P (A, Ω) = P (∅, ∅). Therefore P (∅, ∅) = P (Ω, Ac ). Thus from Theorem 2.2 and Axioms (i), (ii), P (Ω, A) = P (Ω, A) + P (∅, ∅) = P (Ω, Ω) = R. The rest is similarly proved. 2 Theorem 2.6. Let x be any regular element of R. For any A, B ∈ F , if x ∈ P (A, B), then P (A ∩ B, A ∩ B) = R. Proof. Assume x ∈ P (A, B). Then from Theorem 2.2, we have x ∈ P (A, Ω) and x ∈ P (Ω, B). Thus from Lemma 2.1, P (A, Ω) = P (Ω, A) = P (Ω, B) = P (B, Ω) = R. Hence, Theorem 2.4 gives the proof. 2
3. Block calculations Let R be any ring and (Ω, F , P ) be any triplet satisfying Axioms (i)–(iii). In this section, we shall see that each element of R can be divided into block components and block calculations can be applied. Moreover, we shall examine the block decompositions of the
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identity element and show some related theorems concerning ring isomorphisms and generalized inverses. Throughout this section, we assume (i) P (∅, ∅) = {0}. Then we have Theorem 3.1. Let A, B ∈ F . Furthermore, let {Ai ∈ F }ni=1 and {Bj ∈ F }m j =1 be any partitions of A and B, respectively. Then P (A, B) is expressed as a direct sum of abelian groups as follows: P (A, B) =
n m
P (Ai , Bj ).
i=1 j =1
n Proof. From Axioms (i) ,m (ii) and Theorem 2.4, we have P (A, B) = i=1 P (Ai , B). Moreover, P (Ai , B) = j =1 P (Ai , Bj ) (1 i n). Hence we have the proof. 2 Therefore we can give the following definition. Definition 3.1 (Block components). For any x ∈ R and A, B ∈ F , we divide x as x = y11 + y12 + y21 + y22
y11 ∈ P (A, B), y12 ∈ P A, B c , y21 ∈ P Ac , B , y22 ∈ P Ac , B c . Noting that this decomposition is unique, we put xAB = y11 and call it the (A, B)-block of x. The following proposition and theorems are almost obvious. Proposition 3.1. For any A1 , A2 , B1 , B2 ∈ F and x ∈ P (A1 , B1 ) such that A1 ⊂ A2 and B1 ⊂ B2 , xA2 B2 = x,
xA2 B2c = xAc2 B2 = xAc2 B2c = 0.
Especially, xΩΩ = x. Theorem 3.2 (Block decomposition). Let A, B ∈ F . Furthermore let {Ai ∈ F }ni=1 and {Bj ∈ F }m j =1 be any partitions of A and B, respectively. Then for any x ∈ R, xAB =
m n i=1 j =1
xAi Bj .
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Theorem 3.3 (Block calculations: addition). For any x, y ∈ R and A, B ∈ F , (x + y)AB = xAB + yAB . Moreover, we know that block calculations are applicable also to multiplication. Theorem 3.4 (Block calculations: multiplication). Let x, y ∈ R and A, C ∈ F . For any partition {Bi ∈ F }ni=1 of Ω, (xy)AC =
n
xABi yBi C .
i=1
Proof. From Theorem 3.2, we know that x = yΩC c . Thus xy =
n i,j =1
xABi yBj C +
n i=1
n
i=1 xABi
xABi yΩC c +
n
+ xAc Ω and y =
n
j =1 yBj C
+
xAc Ω yBj C + xAc Ω yΩC c .
j =1
From Corollary 2.1, the first, the second, the third, and the fourth term in the right-hand side above belong to P (A, C), P (A, C c ), P (Ac , C), and P (Ac , C c ), respectively. Hence, Definition 3.1 shows that (xy)AC = ni,j =1 xABi yBj C . Furthermore, from Theorem 2.3(c) and Axiom (i) we have ni,j =1 xABi yBj C = ni=1 xABi yBi C , which proves the theorem. 2 The following give the details of the block decompositions of the identity element 1. Lemma 3.1. Let A, B ∈ F and x ∈ P (A, B). Then, 1AA x = x1BB = x. Proof. From Theorem 3.4, xAB = (1 · x)AB = 1AA xAB + 1AAc xAc B . Noting that x ∈ P (A, B), we have from Proposition 3.1 that xAB = x and xAc B = 0. Thus we obtain 1AA x = x. The rest can be proved in the same way. 2 Theorem 3.5. Let A, B, C ∈ F and x ∈ P (A, B). Then, 1CC x = xA∩CB and x1CC = xAB∩C . Proof. Applying Proposition 3.1 and Theorem 3.2, we have x = xAB = xA∩CB + xA\CB . Since xA∩CB ∈ P (C, B), we know from Lemma 3.1 and Theorem 2.3(c) that 1CC x = xA∩CB . The latter part is similarly proved. 2 Theorem 3.6. For any A ∈ F , P (A, A) is a ring with the identity element 1AA . Moreover, for each A, B ∈ F , P (A, B) is a left P (A, A) and right P (B, B) module. Proof. In Theorem 2.1, we have already observed that P (A, A) is an abelian group. Moreover, Corollary 2.1 shows that for any x, y ∈ P (A, A), xy ∈ P (A, A). Furthermore, from
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Lemma 3.1, 1AA x = x1AA = x (x ∈ P (A, A)). The remaining ring axioms are obviously satisfied. Further, from Corollary 2.1 and Lemma 3.1, P (A, B) becomes a bimodule. 2 Lemma 3.2. For any A, B ∈ F such that A ∩ B = ∅, we have 1AB = 0. Proof. From Proposition 3.1, (1AA )AA = 1AA and (1AA )AAc = 0. Thus, noting Theorem 3.4 and Lemma 3.1, we have (1AA 1)AB = (1AA )AA 1AB + (1AA )AAc 1Ac B = 1AB . On the other hand, Proposition 3.1 gives (1AA 1)AB = 0. 2 Corollary 3.1. Let {Ai ∈ F }ni=1 be any partition of A ∈ F . Then
n
i=1 1Ai Ai
= 1AA .
Proof. This follows from Theorem 3.2 and Lemma 3.2. 2 Theorem 3.7. Let A, B ∈ F and x ∈ R. Then, xAB = 1AA x1BB . Therefore, we have {1AA }R{1BB } = P (A, B). Proof. For any x ∈ R, Corollary 3.1 shows x = (1AA + 1Ac Ac )x(1BB + 1B c B c ) = 1AA x1BB + 1AA x1B c B c + 1Ac Ac x1BB + 1Ac Ac x1B c B c . Thus, we obtain from Definition 3.1 that xAB = 1AA x1BB .
2
We shall conclude this section with theorems on ring isomorphisms and generalized inverses. Theorem 3.8. Let R be any ring and (Ω, F , P ) be a triplet that satisfies Axioms (i)–(iii), (i) and the following additional condition: (M) For a partition {Ai ∈ F }ni=1 of Ω, P (Ai , Ω) ∼ = P (A1 , Ω) (2 i n) as right Rmodules. Then, R ∼ = M(n; P (A1 , A1 )) as rings, where M(n; P (A1 , A1 )) denotes the ring that consists of all n × n matrices on P (A1 , A1 ). Proof. The claim is easily verified by using Theorem 3.7 and well-known results. So, we only show the outline of the proof. Let gi be an isomorphism from P (A1 , Ω) to P (Ai , Ω). We put ei1 = gi (1A1 A1 ) and e1i = gi−1 (1Ai Ai ). Then, ei1 ∈ P (Ai , A1 ),
e1i ∈ P (A1 , Ai ),
ei1 e1i = 1Ai ,Ai ,
e1i ei1 = 1A1 ,A1
(1 i n).
For each i, j (1 i, j n) we define a function fij : P (A1 , A1 ) → P (Ai , Aj ) by fij (x) = ei1 xe1j ,
x ∈ P (A1 , A1 ).
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Then, fij is a group isomorphism. Moreover, fij (xy) = fik (x)fkj (y) (1 i, j, k n, x, y ∈ P (A1 , A1 )). Therefore, we can define an isomorphism f : M(n; P (A1, A1 )) → R by f (S) =
n
fij (sij )
S ∈ M n; P (A1 , A1 ) ,
i,j =1
where sij denotes the (i, j )th element of S, that is, S = (sij ).
2
Theorem 3.9 (Generalized inverses). Let R be any ring. For any x, y ∈ R, the following three conditions are equivalent to each other: (a) y is a generalized inverse of x, that is, xyx = x. (b) There exists (Ω, F , P ) satisfying Axioms (i)–(iii) and (i) such that for some A ∈ F , x ∈ P (A, Ω),
(xy)AA = 1AA .
(c) There exists (Ω, F , P ) satisfying Axioms (i)–(iii) and (i) such that for some B ∈ F , x ∈ P (Ω, B),
(yx)BB = 1BB .
Proof. We assume (a) and derive (b). We put n = 2, I3 = {0}, e1 = xy, e2 = 1 − xy and construct (Ω, F , P ) just in the same way as in Example 1.2. Then x = (xy)x ∈ P ({1}, Ω) and xy = 1{1}{1} . Therefore, we have (b). We now derive (a) from (b). From Theorem 3.7, 1AA = (xy)AA = 1AA xy1AA . Therefore, noting x ∈ P (A, Ω), we obtain from Lemma 3.1 that x = 1AA xy1AAx = xyx. Thus (a) and (b) are equivalent. Similarly (a) and (c) are equivalent. 2 Remark 3.1. In the matrix theory, generalized inverses are defined not only for square matrices but also for non-square matrices (see, for example, [1,2]). Let X be any m × n
∈ R by adding zero matrix on C such that m > n. We put R = M(m; C) and define X
= ( X O ). Then, we can apply the above theorem. The same vectors to X, that is, X technique is applicable also for the case where m < n. Theorem 3.10 (Reflexive generalized inverses). Let R be any ring. For any x, y ∈ R, the following two conditions are equivalent to each other: (a) y is a reflexive generalized inverse of x, that is, xyx = x and yxy = y. (b) There exists (Ω, F , P ) satisfying Axioms (i)–(iii) and (i) such that for some A ∈ F , x ∈ P (A, Ω), y ∈ P (Ω, A),
xy = 1AA .
Proof. We assume (a) and define (Ω, F , P ) just in the same way as in the proof of Theorem 3.9. Then y = y(xy) ∈ P (Ω, {1}). Thus we obtain (b). Conversely, (b) implies yxy = y1AA = y because of Lemma 3.1. Hence we have (a). 2
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4. One-parameter case Let R be any ring and (Ω, F , P ) be a triplet satisfying not only (i)–(iii) but also the following axiom: (iv) P (A, Ac ) = P (∅, ∅) (A ∈ F ). We do not assume (i) introduced in Section 3. Example 4.1. Let R be commutative. Then Axiom (iv) is derived from (iii). Proof. For any A ∈ F ,
P A, Ac = P (A, Ω)P Ω, Ac = P Ω, Ac P (A, Ω) = P (∅, ∅). Thus we have (iv). 2 We now prove the following lemma and theorem. Lemma 4.1. If A, B ∈ F , then P (A, B) = P (A ∩ B, A ∩ B). Proof. If A ∩ B = ∅, then Theorem 2.2 implies P (A, B) ⊂ P (A, Ac ) = P (∅, ∅). In general, Axiom (ii) gives P (A, B) = P (A ∩ B, A ∩ B) + P (A \ B, A ∩ B) + P (A ∩ B, B \ A) + P (A \ B, B \ A). We know that the last three terms of the right-hand side above are all equal to P (∅, ∅). Hence we have the proof. 2 Theorem 4.1. For any A, B ∈ F , P (A, B) is a two-sided ideal of R. Proof. From Lemma 4.1, P (A, B) = P (A ∩ B, Ω) = P (Ω, A ∩ B). Hence, Theorem 2.1 gives the proof. 2 We define a one-parameter function Q : F → 2R by Q(A) = P (A, A). Immediately from Lemma 4.1, we obtain the following lemma. Lemma 4.2. For any A, B ∈ F , we have Q(A ∩ B) = P (A, B). Moreover, we can show Theorem 4.2. For any A, B ∈ F ,
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(I) Q(Ω) = R, (II) Q(A) + Q(B) = Q(A ∪ B), (III) Q(A)Q(B) = Q(A ∩ B). Proof. From the definition of Q and Axiom (i), we know that Q(Ω) = P (Ω, Ω) = R. Thus we have (I). With the help of Lemma 4.2 and Axiom (ii), Q(A) + Q(B) = P (A, Ω) + P (B, Ω) = P (A ∪ B, Ω) = Q(A ∪ B), which gives (II). Similarly, (III) follows from Lemma 4.2 and Axiom (iii).
2
Conversely, let Q : F → 2R be any one-parameter function satisfying conditions (I)– (III) in Theorem 4.2. We define a two-parameter function P : F × F → 2R by P (A, B) = Q(A ∩ B). Then we have the following theorem. Theorem 4.3. P satisfies Axioms (i)–(iv). Proof. From (I) we have P (Ω, Ω) = Q(Ω) = R, which gives (i). We know from (II) that for any A, B, C ∈ F ,
P (A, C) + P (B, C) = Q(A ∩ C) + Q(B ∩ C) = Q (A ∪ B) ∩ C = P (A ∪ B, C). Similarly P (A, B) + P (A, C) = P (A, B ∪ C). Thus we have (ii). Condition (iii) is obtained from (III) in the same fashion. Finally P (A, Ac ) = Q(∅) = P (∅, ∅), which gives (iv). 2
5. Functions into the set of all idempotents In this section, we shall characterize Axioms (i)–(iii) and (i) from the viewpoint of idempotents. Let R be any ring and E be the set of all idempotents of R. We introduce a binary relation into E by ef
⇔
ef = f e = e.
Then it is easily verified that the relation becomes a semiorder in E. Let (Ω, F , p) be any triplet such that Ω is a non-empty set, F is a finitely additive class on Ω and p is a function from F to E that satisfies (A) p(Ω) = 1, (B) p(A ∪ B) = p(A) + p(B) (A, B ∈ F , A ∩ B = ∅), (C) p(A) p(B) (A, B ∈ F , A ⊂ B).
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Remark 5.1. If we replace F by a completely additive class, p by a real valued function on F and the finite additivity in (B) by complete additivity, then the above three conditions are equivalent to the Kolmogorov axioms of probability spaces. We shall show the following preliminary results. Lemma 5.1. The function p satisfies (a) (b) (c) (d)
p(∅) = 0, p(A ∪ B) = p(A) + p(B) − p(A ∩ B) (A, B ∈ F ), p(A)p(B) = 0 (A, B ∈ F , A ∩ B = ∅), p(A ∩ B) = p(A)p(B) (A, B ∈ F ).
Proof. From (B), we have p(∅) = p(∅)+p(∅), which gives (a). Moreover, noting A ∪B = (A \ B) ∪ (B \ A) ∪ (A ∩ B), A = (A \ B) ∪ (A ∩ B), and B = (B \ A) ∪ (A ∩ B), we obtain (b) from Axiom (B). To prove (c), let A ∩ B = ∅. Then (B) gives
p(A)p(A ∪ B) = p(A) p(A) + p(B) = p(A) + p(A)p(B). On the other hand, Axiom (C) means p(A)p(A ∪B) = p(A). Thus p(A)p(B) = 0. Finally, we prove (d). From (B), (C), and (c), p(A)p(B) = p(A)p(A ∩ B) + p(A)p(B \ A) = p(A ∩ B). 2 Remark 5.2. We can replace Axiom (C) by (c), that is, (C) is derived from (B) and (c), since for A, B ∈ F such that A ⊂ B, we know that p(A)p(B) = p(A)(p(B \A)+p(A)) = p(A)p(A) = p(A). Similarly, p(B)p(A) = p(A). Thus we obtain (C). Furthermore, if any non-zero element x ∈ R satisfy 2x = 0, then (c) (therefore (C)) is derived from (B). This can be verified as follows. Let A, B ∈ F such that A ∩ B = ∅. Then Axiom (B) implies that p(A) + p(B) is idempotent. Hence
2 p(A) + p(B) = p(A) + p(B) = p(A) + p(A)p(B) + p(B)p(A) + p(B), which means p(A)p(B) = −p(B)p(A). Thus, p(A)p(B) = p(A)p(A)p(B) = p(B)p(A)p(A) = p(B)p(A). Therefore, 2p(A)p(B) = 0. We shall give an example of p that satisfies (A) and (B) but does not satisfy (C). Let R be the ring of all 2 × 2 matrices over the finite field F2 . We put Ω = {1, 2, 3} and F = 2Ω . Moreover, we define a function p by
0 0 1 0 p(∅) = p {1, 3} = , p(Ω) = p {2} = , 0 0 0 1
p {1} = p {3} =
1 0 , 0 0
p {1, 2} = p {2, 3} =
0 0
0 . 1
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Then p satisfies (A) and (B) but does not satisfy (C). For example, p({1})p({1, 2}) = p({1}). We now show that there exists a bijective correspondence between (Ω, F , P ) and (Ω, F , p). Theorem 5.1. Let R be any ring and (Ω, F , P ) be a triplet that satisfies Axioms (i)–(iii) and (i) . We define a function p on F by p(A) = 1AA (A ∈ F ). Then Axioms (A)–(C) hold for (Ω, F , p). Proof. From Proposition 3.1, we have p(Ω) = 1ΩΩ = 1. Moreover, if A ∩ B = ∅, then Corollary 3.1 implies p(A ∪ B) = 1AA + 1BB = p(A) + p(B). Furthermore, if A ⊂ B, then p(A)p(B) = 1AA 1BB = 1AA = p(A) because of Lemma 3.1. Similarly, p(B)p(A) = p(A). Therefore p(A) p(B). 2 Theorem 5.2. Let R be any ring. Moreover, let (Ω, F , p) be a triplet that satisfies Axioms (A)–(C). We define a function P from F × F to 2R by P (A, B) = p(A)Rp(B)
(A, B ∈ F ).
Then Axioms (i)–(iii) and (i) hold for (Ω, F , P ). Moreover, 1AA = p(A). Proof. From Axiom (A), we have P (Ω, Ω) = R. Moreover, Lemma 5.1(a) indicates that P (∅, ∅) = {0}. Thus (i) and (i) hold. Furthermore, from Lemma 5.1(b) and Axiom (C), we have
P (A ∪ B, C) = p(A) + p(B) − p(A ∩ B) Rp(C)
⊂ p(A)Rp(C) + p(B) − p(A ∩ B) Rp(C)
= p(A)Rp(C) + p(B) p(B) − p(A ∩ B) Rp(C) ⊂ p(A)Rp(C) + p(B)Rp(C) = P (A, C) + P (B, C). On the other hand, Axiom (C) gives P (A, C) + P (B, C) = p(A)Rp(C) + p(B)Rp(C) = p(A ∪ B)p(A)Rp(C) + p(A ∪ B)p(B)Rp(C) ⊂ p(A ∪ B)Rp(C) = P (A ∪ B, C). Therefore, P (A, C) + P (B, C) = P (A ∪ B, C). Similarly, we have P (A, B) + P (A, C) = P (A, B ∪C). We now prove (iii). From Axiom (A), P (A, Ω)P (Ω, B) = p(A)RRp(B) =
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P (A, B) and from Lemma 5.1(c), P (Ω, A)P (Ac , Ω) = Rp(A)p(Ac )R = P (∅, ∅). Thus, the triplet (Ω, F , P ) satisfies (i)–(iii) and (i) . Finally, noting Axioms (A) and (B), we obtain 1 = p(A)+p(Ac ). In this decomposition, p(A) ∈ P (A, A) and p(Ac ) ∈ P (Ac , Ac ). Hence we know from Definition 3.1 that 1AA = p(A). 2 In addition, the following theorem gives the detail of the one-parameter case. Theorem 5.3. Let R be any ring and (Ω, F , P ) be a triplet that satisfies not only Axioms (i)–(iii) and (i) but also (iv) in Section 4. We define a function p on F by p(A) = 1AA (A ∈ F ). Then, for each A ∈ F , p(A) is in the center of R. Conversely, let (Ω, F , p) be a triplet that satisfies Axioms (A)–(C). We also assume that p is a function into the set of all central idempotents of R and define a function P from F × F to 2R by P (A, B) = p(A)Rp(B)
(A, B ∈ F ).
Then (Ω, F , P ) satisfies (iv). Proof. Let (Ω, F , P ) satisfies (iv). We first note that if A ∈ F , then for each x ∈ R, 1AA x1Ac Ac = 1Ac Ac x1AA = 0 since 1AA x1Ac Ac ∈ P (A, Ac ) and 1Ac Ac x1AA ∈ P (Ac , A). Therefore, from Corollary 3.1, we have 1AA x = 1AA x(1AA + 1Ac Ac ) = 1AA x1AA and x1AA = (1AA + 1Ac Ac )x1AA = 1AA x1AA . Hence 1AA belongs to the center of R. We shall show the converse. Let A ∈ F . Then, from Lemma 5.1(c), P (A, Ac ) = p(A)p(Ac )R = {0}. This gives the proof. 2 Especially, focusing our attention on a partition {Ai ∈ F }ni=1 of Ω, we know that the direct product of rings P (Ai , Ai ), the direct sum of two-sided ideals P (Ai , Ai ) and the central idempotent decomposition 1 = ni=1 1Ai Ai correspond to each other. This fact has been formulated in [4]. In particular, block decompositions associated with central primitive idempotent decompositions play important roles in the representation theory of finite groups [3].
References [1] A. Ben-Israel, T.N.E. Greville, Generalized Inverses, Wiley, New York, 1974. [2] K.P.S. Bhaskara Rao, The Theory of Generalized Inverses Over Commutative Rings, Taylor & Francis, New York, 2002. [3] W. Feit, The Representation Theory of Finite Groups, North-Holland, Amsterdam, 1982. [4] B. Pierce, Linear associative algebra, Amer. J. Math. 4 (1881) 97–229.
An axiomatic approach to block decompositions of rings Masaya Matsuura Department of Mathematical Informatics, Graduate School of Information Science and Technology, University of Tokyo, Bunkyo-ku, Tokyo 113-8656, Japan Received 5 February 2004 Available online 13 December 2004 Communicated by Paul Flavell
Abstract In calculations with matrices, block calculations play an important role. To elucidate the essential structures of block decompositions, we shall in this paper introduce a simple system of axioms which guarantees block calculations of rings. The axioms can be interpreted as rules of some kind of information filters. We shall also give another system of axioms in terms of idempotents which is equivalent to the above mentioned system and is similar to the Kolmogorov axioms of probability spaces. 2004 Elsevier Inc. All rights reserved. Keywords: Block decompositions; Block calculations; Idempotents
1. Introduction We assume that each ring treated in this paper has an identity element 1. Let R be any ring and (Ω, F , P ) be any triplet such that Ω is a non-empty set, F is a finitely additive class on Ω and P is a function from F × F to 2R that satisfies (i) P (Ω, Ω) = R, E-mail address: [email protected]. 0021-8693/$ – see front matter 2004 Elsevier Inc. All rights reserved. doi:10.1016/j.jalgebra.2004.10.012
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(ii) P (A, C) + P (B, C) = P (A ∪ B, C), P (A, B) + P (A, C) = P (A, B ∪ C), (iii) P (A, Ω)P (Ω, B) = P (A, B), P (Ω, A)P (Ac , Ω) = P (∅, ∅), for any A, B, C ∈ F . Here the sets S + T and ST (S, T ∈ 2R ) are defined by S + T = {s + t; s ∈ S, t ∈ T },
ST = {st; s ∈ S, t ∈ T }.
If S = ∅ or T = ∅, then we put S + T = ST = ∅. The triplet (Ω, F , P ) can be seen as an information filter with an entrance and an exit. The ring R corresponds to an information set. The first and the second variable of the function P correspond to the entrance and the exit of the filter, respectively. Each element of F indicates the situation of the filter. For example, P (Ω, ∅) indicates the elements of R which can go through this filter when the entrance of the filter is perfectly open and the exit is completely closed. More generally, P (A, B) indicates the information which can pass the filter when the situations of the entrance and the exit are A and B, respectively. Axiom (i) indicates that if both the entrance and the exit are completely open, then all information can go through the filter. We note that in general, even if both the entrance and the exit are completely closed, some information can pass this filter. Axiom (ii) gives a rule of parallel connections of the filter. Axiom (iii) corresponds to series connections. In Section 2, we shall investigate fundamental properties of (Ω, F , P ). In Section 3, we shall see that under an additional condition, each element of R can be divided into block components and block calculations can be applied. Moreover, generalized inverses are examined in connection with (Ω, F , P ). Section 4 is devoted to the study of a special case where the function P is reduced to a one-parameter function. In Section 5, we shall characterize (Ω, F , P ) in terms of idempotents of R. The argument in this paper can be interpreted as reformulation of classical topics on decompositions of rings from a new axiomatic point of view (cf. [3,4]). We conclude this section with two examples. Example 1.1. Let R = ni=1 Ii be a ring such that Ii (i = 1, 2, . . . , n) is a two-sided ideal of R. Then, we can construct (Ω, F , P ) that satisfies Axioms (i)–(iii) as follows: Ω = {1, 2, . . . , n}, P (A, B) = {0} (A ∩ B = ∅),
F = 2Ω ,
P (A, B) =
i∈A∩B
This triplet has the following additional properties: (i) P (∅, ∅) = {0}, (iv) P (A, Ac ) = P (∅, ∅) (A ∈ F ).
Ii
(otherwise).
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Example 1.2. More generally, let R = n+1 i=1 Ii be a ring such that Ii (i = 1, 2, . . . , n + 1) is a right ideal of R. Further, we assume I n+1 is a two-sided ideal of R. Then, there exists an idempotent decomposition 1 = n+1 i=1 ei such that ei ej = 0
(i = j ),
Ii = ei R (1 i n + 1).
We define (Ω, F , P ) as Ω = {1, 2, . . . , n}, F = 2Ω , ei R ei + In+1 (A, B ∈ F ). P (A, B) = i∈A
i∈B
Then, (Ω, F , P ) satisfies Axioms (i)–(iii). Especially, P ({i}, Ω) = Ii ⊕ In+1 (1 i n) and P (∅, ∅) = In+1 . For example, we can show P (A, Ω)P (Ω, B) = P (A, B) as follows. Since In+1 = en+1 R is a two-sided ideal, we have Ren+1 R = en+1 R. Therefore ei Ren+1 = {0}, ei R ei = ei R. i∈A
i∈A
i∈Ω
i∈A
So ei R ei R ei + In+1 = P (A, B). P (A, Ω)P (Ω, B) ⊂ i∈A
i∈Ω
i∈B
To show the converse relation, let x ∈ P (A, B). Then, there exist r ∈ R and s ∈ In+1 such that x= ei r ei + s. i∈A
i∈B
Hence, noting that s ∈ en+1 R, we have ei r ei + en+1 ei · 1 · ei + s = x. i∈A
i∈Ω
i∈Ω
i∈B
This gives P (A, B) ⊂ P (A, Ω)P (Ω, B). Other axioms can be verified in a similar fashion.
2. Fundamental properties Let R be any ring. In this section, we shall investigate fundamental properties of any triplet (Ω, F , P ) satisfying Axioms (i)–(iii). First, we shall see that P (A, B) is a subgroup of R.
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Theorem 2.1. For any A, B ∈ F , P (A, B) is a subgroup of R. Moreover, P (A, Ω) and P (Ω, B) are a right and a left ideal of R, respectively. Proof. Let A, B ∈ F . Then from Axioms (i) and (ii),
P (A, B) + P A, B c + P Ac , B + P Ac , B c = R. Thus, P (A, B) is non-empty. Let x ∈ P (A, B). Then Axiom (iii) shows there exist y ∈ P (A, Ω) and z ∈ P (Ω, B) such that x = yz. Furthermore, Axiom (i) implies that −1 ∈ P (Ω, Ω). Hence from Axiom (iii), we have y · (−1) ∈ P (A, Ω)P (Ω, Ω) = P (A, Ω). So, noting Axiom (iii) again, we have −x = (−y) · z ∈ P (A, Ω)P (Ω, B) = P (A, B). Moreover, Axiom (ii) implies that if x, y ∈ P (A, B), then x + y ∈ P (A, B). Hence P (A, B) is a subgroup. Furthermore, from Axioms (i) and (iii), P (A, Ω)R = P (A, Ω) and RP (Ω, B) = P (Ω, B). This concludes the proof. 2 Theorem 2.2. Let A1 , A2 , B1 , B2 ∈ F satisfy A1 ⊂ A2 and B1 ⊂ B2 . Then, P (A1 , B1 ) ⊂ P (A2 , B2 ), that is, P (A1 , B1 ) is a subgroup of P (A2 , B2 ). Proof. With the help of Axiom (ii), we have P (A1 , B1 ) + P (A2 , B1 ) = P (A2 , B1 ). From Theorem 2.1, 0 ∈ P (A2 , B1 ). Thus P (A1 , B1 ) ⊂ P (A2 , B1 ). Similarly P (A2 , B1 ) ⊂ P (A2 , B2 ). 2 Corollary 2.1. For any A1 , A2 , B1 , B2 ∈ F , we have P (A1 , B1 )P (A2 , B2 ) ⊂ P (A1 , B2 ). Proof. This follows from Theorem 2.2 and Axiom (iii).
2
The next aim is to examine P (∅, ∅). Theorem 2.3. The group P (∅, ∅) satisfies (a) (b) (c) (d)
P (A, ∅) = P (∅, A) = P (∅, ∅) (A ∈ F ), P (∅, ∅)P (∅, ∅) = P (∅, ∅), P (A, B1 )P (B2 , C) = P (∅, ∅) (A, B1 , B2 , C ∈ F , B1 ∩ B2 = ∅), P (∅, ∅) is a two-sided ideal of R.
Proof. Let A ∈ F . From Theorem 2.2, P (∅, ∅) ⊂ P (A, ∅). Further, from Axioms (i), (iii) and Theorem 2.2, P (A, ∅) ⊂ P (A, ∅)P (Ω, Ω) ⊂ P (Ω, ∅)P (Ω, Ω) = P (∅, ∅).
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Thus P (A, ∅) = P (∅, ∅). Similarly, P (∅, A) = P (∅, ∅). Therefore, from Axiom (iii), P (∅, ∅)P (∅, ∅) = P (∅, Ω)P (Ω, ∅) = P (∅, ∅). To show (c), let A, B1 , B2 , C ∈ F and B1 ∩ B2 = ∅. From Theorem 2.2 and Axiom (iii),
P (A, B1 )P (B2 , C) ⊂ P (Ω, B1 )P B1c , Ω = P (∅, ∅). Furthermore, from Theorem 2.2, P (A, B1 )P (B2 , C) ⊃ P (∅, ∅)P (∅, ∅) = P (∅, ∅). Thus we have (c). So, P (∅, ∅)R = P (∅, ∅)P (Ω, Ω) = P (∅, ∅). Similarly, RP (∅, ∅) = P (∅, ∅). This gives (d). 2 Theorem 2.4. For any A1 , A2 , B1 , B2 ∈ F , P (A1 , B1 ) ∩ P (A2 , B2 ) = P (A1 ∩ A2 , B1 ∩ B2 ). In particular, if A1 ∩ A2 = ∅ or B1 ∩ B2 = ∅, then P (A1 , B1 ) ∩ P (A2 , B2 ) = P (∅, ∅). Proof. From Theorem 2.2, we have P (A1 ∩ A2 , B1 ) ∩ P (A1 ∩ A2 , B2 ) ⊂ P (A1 , B1 ) ∩ P (A2 , B2 ). We shall show the converse relation. Let x ∈ P (A1 , B1 ) ∩ P (A2 , B2 ). From Axioms (i) and (ii), 1 and x are divided as 1 = e1 + e2 + e3 , x = x1 + x2 ,
e1 ∈ P (Ω, A1 ∩ A2 ), e2 ∈ P Ω, Ac1 , e3 ∈ P Ω, Ac2 ,
x1 ∈ P (A1 ∩ A2 , B1 ), x2 ∈ P (A1 \ A2 , B1 ).
Then, x = e1 x + (e2 + e3 )x = e1 x1 + e1 x2 + (e2 + e3 )x = x1 − (e2 + e3 )x1 + e1 x2 + (e2 + e3 )x. From Theorem 2.3(c), we know that all terms except the first term of the last line of the above equation belong to P (∅, ∅). Therefore, Theorem 2.2 gives x ∈ P (A1 ∩ A2 , B1 ). Similarly, x ∈ P (A1 ∩ A2 , B2 ). Thus, P (A1 , B1 ) ∩ P (A2 , B2 ) = P (A1 ∩ A2 , B1 ) ∩ P (A1 ∩ A2 , B2 ). Repeating a similar procedure, we obtain P (A1 ∩ A2 , B1 ) ∩ P (A1 ∩ A2 , B2 ) = P (A1 ∩ A2 , B1 ∩ B2 ).
2
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Concerning the quotient ring R/P (∅, ∅), we have Theorem 2.5. Let R be a quotient ring of R defined by R = R/P (∅, ∅). We denote x = x + P (∅, ∅) (x ∈ R). Furthermore, we define a function P : F × F → 2R by P (A, B) = x; x ∈ P (A, B)
(A, B ∈ F ).
Then P is well-defined and has the following properties: (a) The triplet (Ω, F , P ) satisfies Axioms (i)–(iii) for R. (b) Let A, B ∈ F . Then P (A, B) = R if and only if P (A, B) = R. (c) Let A, B ∈ F . Then P (A, B) = {0} if and only if P (A, B) = P (∅, ∅). Especially, P (∅, ∅) = {0}. Proof. Since it is a routine work to show (a)–(c), we only verify that the function P is well-defined. Let x ∈ P (A, B) and x = y. Then from Theorem 2.2, y = x + (y − x) ∈ P (A, B) + P (∅, ∅) = P (A, B). Thus, P is well-defined. 2 At the end of this section, we shall investigate when P (A, B) becomes R. Lemma 2.1. Let x be any regular element of R. For any A ∈ F , if x ∈ P (A, Ω)∪P (Ω, A), then P (A, Ω) = P (Ω, A) = R. Proof. Assume x ∈ P (A, Ω). From Theorem 2.1, P (A, Ω) is a right ideal. Hence, P (A, Ω) = R. We now show P (Ω, A) = R. From Axiom (iii), P (Ω, Ac ) ⊂ P (Ω, Ac ) × P (A, Ω) = P (∅, ∅). Therefore P (∅, ∅) = P (Ω, Ac ). Thus from Theorem 2.2 and Axioms (i), (ii), P (Ω, A) = P (Ω, A) + P (∅, ∅) = P (Ω, Ω) = R. The rest is similarly proved. 2 Theorem 2.6. Let x be any regular element of R. For any A, B ∈ F , if x ∈ P (A, B), then P (A ∩ B, A ∩ B) = R. Proof. Assume x ∈ P (A, B). Then from Theorem 2.2, we have x ∈ P (A, Ω) and x ∈ P (Ω, B). Thus from Lemma 2.1, P (A, Ω) = P (Ω, A) = P (Ω, B) = P (B, Ω) = R. Hence, Theorem 2.4 gives the proof. 2
3. Block calculations Let R be any ring and (Ω, F , P ) be any triplet satisfying Axioms (i)–(iii). In this section, we shall see that each element of R can be divided into block components and block calculations can be applied. Moreover, we shall examine the block decompositions of the
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identity element and show some related theorems concerning ring isomorphisms and generalized inverses. Throughout this section, we assume (i) P (∅, ∅) = {0}. Then we have Theorem 3.1. Let A, B ∈ F . Furthermore, let {Ai ∈ F }ni=1 and {Bj ∈ F }m j =1 be any partitions of A and B, respectively. Then P (A, B) is expressed as a direct sum of abelian groups as follows: P (A, B) =
n m
P (Ai , Bj ).
i=1 j =1
n Proof. From Axioms (i) ,m (ii) and Theorem 2.4, we have P (A, B) = i=1 P (Ai , B). Moreover, P (Ai , B) = j =1 P (Ai , Bj ) (1 i n). Hence we have the proof. 2 Therefore we can give the following definition. Definition 3.1 (Block components). For any x ∈ R and A, B ∈ F , we divide x as x = y11 + y12 + y21 + y22
y11 ∈ P (A, B), y12 ∈ P A, B c , y21 ∈ P Ac , B , y22 ∈ P Ac , B c . Noting that this decomposition is unique, we put xAB = y11 and call it the (A, B)-block of x. The following proposition and theorems are almost obvious. Proposition 3.1. For any A1 , A2 , B1 , B2 ∈ F and x ∈ P (A1 , B1 ) such that A1 ⊂ A2 and B1 ⊂ B2 , xA2 B2 = x,
xA2 B2c = xAc2 B2 = xAc2 B2c = 0.
Especially, xΩΩ = x. Theorem 3.2 (Block decomposition). Let A, B ∈ F . Furthermore let {Ai ∈ F }ni=1 and {Bj ∈ F }m j =1 be any partitions of A and B, respectively. Then for any x ∈ R, xAB =
m n i=1 j =1
xAi Bj .
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Theorem 3.3 (Block calculations: addition). For any x, y ∈ R and A, B ∈ F , (x + y)AB = xAB + yAB . Moreover, we know that block calculations are applicable also to multiplication. Theorem 3.4 (Block calculations: multiplication). Let x, y ∈ R and A, C ∈ F . For any partition {Bi ∈ F }ni=1 of Ω, (xy)AC =
n
xABi yBi C .
i=1
Proof. From Theorem 3.2, we know that x = yΩC c . Thus xy =
n i,j =1
xABi yBj C +
n i=1
n
i=1 xABi
xABi yΩC c +
n
+ xAc Ω and y =
n
j =1 yBj C
+
xAc Ω yBj C + xAc Ω yΩC c .
j =1
From Corollary 2.1, the first, the second, the third, and the fourth term in the right-hand side above belong to P (A, C), P (A, C c ), P (Ac , C), and P (Ac , C c ), respectively. Hence, Definition 3.1 shows that (xy)AC = ni,j =1 xABi yBj C . Furthermore, from Theorem 2.3(c) and Axiom (i) we have ni,j =1 xABi yBj C = ni=1 xABi yBi C , which proves the theorem. 2 The following give the details of the block decompositions of the identity element 1. Lemma 3.1. Let A, B ∈ F and x ∈ P (A, B). Then, 1AA x = x1BB = x. Proof. From Theorem 3.4, xAB = (1 · x)AB = 1AA xAB + 1AAc xAc B . Noting that x ∈ P (A, B), we have from Proposition 3.1 that xAB = x and xAc B = 0. Thus we obtain 1AA x = x. The rest can be proved in the same way. 2 Theorem 3.5. Let A, B, C ∈ F and x ∈ P (A, B). Then, 1CC x = xA∩CB and x1CC = xAB∩C . Proof. Applying Proposition 3.1 and Theorem 3.2, we have x = xAB = xA∩CB + xA\CB . Since xA∩CB ∈ P (C, B), we know from Lemma 3.1 and Theorem 2.3(c) that 1CC x = xA∩CB . The latter part is similarly proved. 2 Theorem 3.6. For any A ∈ F , P (A, A) is a ring with the identity element 1AA . Moreover, for each A, B ∈ F , P (A, B) is a left P (A, A) and right P (B, B) module. Proof. In Theorem 2.1, we have already observed that P (A, A) is an abelian group. Moreover, Corollary 2.1 shows that for any x, y ∈ P (A, A), xy ∈ P (A, A). Furthermore, from
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Lemma 3.1, 1AA x = x1AA = x (x ∈ P (A, A)). The remaining ring axioms are obviously satisfied. Further, from Corollary 2.1 and Lemma 3.1, P (A, B) becomes a bimodule. 2 Lemma 3.2. For any A, B ∈ F such that A ∩ B = ∅, we have 1AB = 0. Proof. From Proposition 3.1, (1AA )AA = 1AA and (1AA )AAc = 0. Thus, noting Theorem 3.4 and Lemma 3.1, we have (1AA 1)AB = (1AA )AA 1AB + (1AA )AAc 1Ac B = 1AB . On the other hand, Proposition 3.1 gives (1AA 1)AB = 0. 2 Corollary 3.1. Let {Ai ∈ F }ni=1 be any partition of A ∈ F . Then
n
i=1 1Ai Ai
= 1AA .
Proof. This follows from Theorem 3.2 and Lemma 3.2. 2 Theorem 3.7. Let A, B ∈ F and x ∈ R. Then, xAB = 1AA x1BB . Therefore, we have {1AA }R{1BB } = P (A, B). Proof. For any x ∈ R, Corollary 3.1 shows x = (1AA + 1Ac Ac )x(1BB + 1B c B c ) = 1AA x1BB + 1AA x1B c B c + 1Ac Ac x1BB + 1Ac Ac x1B c B c . Thus, we obtain from Definition 3.1 that xAB = 1AA x1BB .
2
We shall conclude this section with theorems on ring isomorphisms and generalized inverses. Theorem 3.8. Let R be any ring and (Ω, F , P ) be a triplet that satisfies Axioms (i)–(iii), (i) and the following additional condition: (M) For a partition {Ai ∈ F }ni=1 of Ω, P (Ai , Ω) ∼ = P (A1 , Ω) (2 i n) as right Rmodules. Then, R ∼ = M(n; P (A1 , A1 )) as rings, where M(n; P (A1 , A1 )) denotes the ring that consists of all n × n matrices on P (A1 , A1 ). Proof. The claim is easily verified by using Theorem 3.7 and well-known results. So, we only show the outline of the proof. Let gi be an isomorphism from P (A1 , Ω) to P (Ai , Ω). We put ei1 = gi (1A1 A1 ) and e1i = gi−1 (1Ai Ai ). Then, ei1 ∈ P (Ai , A1 ),
e1i ∈ P (A1 , Ai ),
ei1 e1i = 1Ai ,Ai ,
e1i ei1 = 1A1 ,A1
(1 i n).
For each i, j (1 i, j n) we define a function fij : P (A1 , A1 ) → P (Ai , Aj ) by fij (x) = ei1 xe1j ,
x ∈ P (A1 , A1 ).
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Then, fij is a group isomorphism. Moreover, fij (xy) = fik (x)fkj (y) (1 i, j, k n, x, y ∈ P (A1 , A1 )). Therefore, we can define an isomorphism f : M(n; P (A1, A1 )) → R by f (S) =
n
fij (sij )
S ∈ M n; P (A1 , A1 ) ,
i,j =1
where sij denotes the (i, j )th element of S, that is, S = (sij ).
2
Theorem 3.9 (Generalized inverses). Let R be any ring. For any x, y ∈ R, the following three conditions are equivalent to each other: (a) y is a generalized inverse of x, that is, xyx = x. (b) There exists (Ω, F , P ) satisfying Axioms (i)–(iii) and (i) such that for some A ∈ F , x ∈ P (A, Ω),
(xy)AA = 1AA .
(c) There exists (Ω, F , P ) satisfying Axioms (i)–(iii) and (i) such that for some B ∈ F , x ∈ P (Ω, B),
(yx)BB = 1BB .
Proof. We assume (a) and derive (b). We put n = 2, I3 = {0}, e1 = xy, e2 = 1 − xy and construct (Ω, F , P ) just in the same way as in Example 1.2. Then x = (xy)x ∈ P ({1}, Ω) and xy = 1{1}{1} . Therefore, we have (b). We now derive (a) from (b). From Theorem 3.7, 1AA = (xy)AA = 1AA xy1AA . Therefore, noting x ∈ P (A, Ω), we obtain from Lemma 3.1 that x = 1AA xy1AAx = xyx. Thus (a) and (b) are equivalent. Similarly (a) and (c) are equivalent. 2 Remark 3.1. In the matrix theory, generalized inverses are defined not only for square matrices but also for non-square matrices (see, for example, [1,2]). Let X be any m × n
∈ R by adding zero matrix on C such that m > n. We put R = M(m; C) and define X
= ( X O ). Then, we can apply the above theorem. The same vectors to X, that is, X technique is applicable also for the case where m < n. Theorem 3.10 (Reflexive generalized inverses). Let R be any ring. For any x, y ∈ R, the following two conditions are equivalent to each other: (a) y is a reflexive generalized inverse of x, that is, xyx = x and yxy = y. (b) There exists (Ω, F , P ) satisfying Axioms (i)–(iii) and (i) such that for some A ∈ F , x ∈ P (A, Ω), y ∈ P (Ω, A),
xy = 1AA .
Proof. We assume (a) and define (Ω, F , P ) just in the same way as in the proof of Theorem 3.9. Then y = y(xy) ∈ P (Ω, {1}). Thus we obtain (b). Conversely, (b) implies yxy = y1AA = y because of Lemma 3.1. Hence we have (a). 2
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4. One-parameter case Let R be any ring and (Ω, F , P ) be a triplet satisfying not only (i)–(iii) but also the following axiom: (iv) P (A, Ac ) = P (∅, ∅) (A ∈ F ). We do not assume (i) introduced in Section 3. Example 4.1. Let R be commutative. Then Axiom (iv) is derived from (iii). Proof. For any A ∈ F ,
P A, Ac = P (A, Ω)P Ω, Ac = P Ω, Ac P (A, Ω) = P (∅, ∅). Thus we have (iv). 2 We now prove the following lemma and theorem. Lemma 4.1. If A, B ∈ F , then P (A, B) = P (A ∩ B, A ∩ B). Proof. If A ∩ B = ∅, then Theorem 2.2 implies P (A, B) ⊂ P (A, Ac ) = P (∅, ∅). In general, Axiom (ii) gives P (A, B) = P (A ∩ B, A ∩ B) + P (A \ B, A ∩ B) + P (A ∩ B, B \ A) + P (A \ B, B \ A). We know that the last three terms of the right-hand side above are all equal to P (∅, ∅). Hence we have the proof. 2 Theorem 4.1. For any A, B ∈ F , P (A, B) is a two-sided ideal of R. Proof. From Lemma 4.1, P (A, B) = P (A ∩ B, Ω) = P (Ω, A ∩ B). Hence, Theorem 2.1 gives the proof. 2 We define a one-parameter function Q : F → 2R by Q(A) = P (A, A). Immediately from Lemma 4.1, we obtain the following lemma. Lemma 4.2. For any A, B ∈ F , we have Q(A ∩ B) = P (A, B). Moreover, we can show Theorem 4.2. For any A, B ∈ F ,
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(I) Q(Ω) = R, (II) Q(A) + Q(B) = Q(A ∪ B), (III) Q(A)Q(B) = Q(A ∩ B). Proof. From the definition of Q and Axiom (i), we know that Q(Ω) = P (Ω, Ω) = R. Thus we have (I). With the help of Lemma 4.2 and Axiom (ii), Q(A) + Q(B) = P (A, Ω) + P (B, Ω) = P (A ∪ B, Ω) = Q(A ∪ B), which gives (II). Similarly, (III) follows from Lemma 4.2 and Axiom (iii).
2
Conversely, let Q : F → 2R be any one-parameter function satisfying conditions (I)– (III) in Theorem 4.2. We define a two-parameter function P : F × F → 2R by P (A, B) = Q(A ∩ B). Then we have the following theorem. Theorem 4.3. P satisfies Axioms (i)–(iv). Proof. From (I) we have P (Ω, Ω) = Q(Ω) = R, which gives (i). We know from (II) that for any A, B, C ∈ F ,
P (A, C) + P (B, C) = Q(A ∩ C) + Q(B ∩ C) = Q (A ∪ B) ∩ C = P (A ∪ B, C). Similarly P (A, B) + P (A, C) = P (A, B ∪ C). Thus we have (ii). Condition (iii) is obtained from (III) in the same fashion. Finally P (A, Ac ) = Q(∅) = P (∅, ∅), which gives (iv). 2
5. Functions into the set of all idempotents In this section, we shall characterize Axioms (i)–(iii) and (i) from the viewpoint of idempotents. Let R be any ring and E be the set of all idempotents of R. We introduce a binary relation into E by ef
⇔
ef = f e = e.
Then it is easily verified that the relation becomes a semiorder in E. Let (Ω, F , p) be any triplet such that Ω is a non-empty set, F is a finitely additive class on Ω and p is a function from F to E that satisfies (A) p(Ω) = 1, (B) p(A ∪ B) = p(A) + p(B) (A, B ∈ F , A ∩ B = ∅), (C) p(A) p(B) (A, B ∈ F , A ⊂ B).
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Remark 5.1. If we replace F by a completely additive class, p by a real valued function on F and the finite additivity in (B) by complete additivity, then the above three conditions are equivalent to the Kolmogorov axioms of probability spaces. We shall show the following preliminary results. Lemma 5.1. The function p satisfies (a) (b) (c) (d)
p(∅) = 0, p(A ∪ B) = p(A) + p(B) − p(A ∩ B) (A, B ∈ F ), p(A)p(B) = 0 (A, B ∈ F , A ∩ B = ∅), p(A ∩ B) = p(A)p(B) (A, B ∈ F ).
Proof. From (B), we have p(∅) = p(∅)+p(∅), which gives (a). Moreover, noting A ∪B = (A \ B) ∪ (B \ A) ∪ (A ∩ B), A = (A \ B) ∪ (A ∩ B), and B = (B \ A) ∪ (A ∩ B), we obtain (b) from Axiom (B). To prove (c), let A ∩ B = ∅. Then (B) gives
p(A)p(A ∪ B) = p(A) p(A) + p(B) = p(A) + p(A)p(B). On the other hand, Axiom (C) means p(A)p(A ∪B) = p(A). Thus p(A)p(B) = 0. Finally, we prove (d). From (B), (C), and (c), p(A)p(B) = p(A)p(A ∩ B) + p(A)p(B \ A) = p(A ∩ B). 2 Remark 5.2. We can replace Axiom (C) by (c), that is, (C) is derived from (B) and (c), since for A, B ∈ F such that A ⊂ B, we know that p(A)p(B) = p(A)(p(B \A)+p(A)) = p(A)p(A) = p(A). Similarly, p(B)p(A) = p(A). Thus we obtain (C). Furthermore, if any non-zero element x ∈ R satisfy 2x = 0, then (c) (therefore (C)) is derived from (B). This can be verified as follows. Let A, B ∈ F such that A ∩ B = ∅. Then Axiom (B) implies that p(A) + p(B) is idempotent. Hence
2 p(A) + p(B) = p(A) + p(B) = p(A) + p(A)p(B) + p(B)p(A) + p(B), which means p(A)p(B) = −p(B)p(A). Thus, p(A)p(B) = p(A)p(A)p(B) = p(B)p(A)p(A) = p(B)p(A). Therefore, 2p(A)p(B) = 0. We shall give an example of p that satisfies (A) and (B) but does not satisfy (C). Let R be the ring of all 2 × 2 matrices over the finite field F2 . We put Ω = {1, 2, 3} and F = 2Ω . Moreover, we define a function p by
0 0 1 0 p(∅) = p {1, 3} = , p(Ω) = p {2} = , 0 0 0 1
p {1} = p {3} =
1 0 , 0 0
p {1, 2} = p {2, 3} =
0 0
0 . 1
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Then p satisfies (A) and (B) but does not satisfy (C). For example, p({1})p({1, 2}) = p({1}). We now show that there exists a bijective correspondence between (Ω, F , P ) and (Ω, F , p). Theorem 5.1. Let R be any ring and (Ω, F , P ) be a triplet that satisfies Axioms (i)–(iii) and (i) . We define a function p on F by p(A) = 1AA (A ∈ F ). Then Axioms (A)–(C) hold for (Ω, F , p). Proof. From Proposition 3.1, we have p(Ω) = 1ΩΩ = 1. Moreover, if A ∩ B = ∅, then Corollary 3.1 implies p(A ∪ B) = 1AA + 1BB = p(A) + p(B). Furthermore, if A ⊂ B, then p(A)p(B) = 1AA 1BB = 1AA = p(A) because of Lemma 3.1. Similarly, p(B)p(A) = p(A). Therefore p(A) p(B). 2 Theorem 5.2. Let R be any ring. Moreover, let (Ω, F , p) be a triplet that satisfies Axioms (A)–(C). We define a function P from F × F to 2R by P (A, B) = p(A)Rp(B)
(A, B ∈ F ).
Then Axioms (i)–(iii) and (i) hold for (Ω, F , P ). Moreover, 1AA = p(A). Proof. From Axiom (A), we have P (Ω, Ω) = R. Moreover, Lemma 5.1(a) indicates that P (∅, ∅) = {0}. Thus (i) and (i) hold. Furthermore, from Lemma 5.1(b) and Axiom (C), we have
P (A ∪ B, C) = p(A) + p(B) − p(A ∩ B) Rp(C)
⊂ p(A)Rp(C) + p(B) − p(A ∩ B) Rp(C)
= p(A)Rp(C) + p(B) p(B) − p(A ∩ B) Rp(C) ⊂ p(A)Rp(C) + p(B)Rp(C) = P (A, C) + P (B, C). On the other hand, Axiom (C) gives P (A, C) + P (B, C) = p(A)Rp(C) + p(B)Rp(C) = p(A ∪ B)p(A)Rp(C) + p(A ∪ B)p(B)Rp(C) ⊂ p(A ∪ B)Rp(C) = P (A ∪ B, C). Therefore, P (A, C) + P (B, C) = P (A ∪ B, C). Similarly, we have P (A, B) + P (A, C) = P (A, B ∪C). We now prove (iii). From Axiom (A), P (A, Ω)P (Ω, B) = p(A)RRp(B) =
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P (A, B) and from Lemma 5.1(c), P (Ω, A)P (Ac , Ω) = Rp(A)p(Ac )R = P (∅, ∅). Thus, the triplet (Ω, F , P ) satisfies (i)–(iii) and (i) . Finally, noting Axioms (A) and (B), we obtain 1 = p(A)+p(Ac ). In this decomposition, p(A) ∈ P (A, A) and p(Ac ) ∈ P (Ac , Ac ). Hence we know from Definition 3.1 that 1AA = p(A). 2 In addition, the following theorem gives the detail of the one-parameter case. Theorem 5.3. Let R be any ring and (Ω, F , P ) be a triplet that satisfies not only Axioms (i)–(iii) and (i) but also (iv) in Section 4. We define a function p on F by p(A) = 1AA (A ∈ F ). Then, for each A ∈ F , p(A) is in the center of R. Conversely, let (Ω, F , p) be a triplet that satisfies Axioms (A)–(C). We also assume that p is a function into the set of all central idempotents of R and define a function P from F × F to 2R by P (A, B) = p(A)Rp(B)
(A, B ∈ F ).
Then (Ω, F , P ) satisfies (iv). Proof. Let (Ω, F , P ) satisfies (iv). We first note that if A ∈ F , then for each x ∈ R, 1AA x1Ac Ac = 1Ac Ac x1AA = 0 since 1AA x1Ac Ac ∈ P (A, Ac ) and 1Ac Ac x1AA ∈ P (Ac , A). Therefore, from Corollary 3.1, we have 1AA x = 1AA x(1AA + 1Ac Ac ) = 1AA x1AA and x1AA = (1AA + 1Ac Ac )x1AA = 1AA x1AA . Hence 1AA belongs to the center of R. We shall show the converse. Let A ∈ F . Then, from Lemma 5.1(c), P (A, Ac ) = p(A)p(Ac )R = {0}. This gives the proof. 2 Especially, focusing our attention on a partition {Ai ∈ F }ni=1 of Ω, we know that the direct product of rings P (Ai , Ai ), the direct sum of two-sided ideals P (Ai , Ai ) and the central idempotent decomposition 1 = ni=1 1Ai Ai correspond to each other. This fact has been formulated in [4]. In particular, block decompositions associated with central primitive idempotent decompositions play important roles in the representation theory of finite groups [3].
References [1] A. Ben-Israel, T.N.E. Greville, Generalized Inverses, Wiley, New York, 1974. [2] K.P.S. Bhaskara Rao, The Theory of Generalized Inverses Over Commutative Rings, Taylor & Francis, New York, 2002. [3] W. Feit, The Representation Theory of Finite Groups, North-Holland, Amsterdam, 1982. [4] B. Pierce, Linear associative algebra, Amer. J. Math. 4 (1881) 97–229.